| If \[\left( a+\frac{1}{a} \right)=4\sqrt{2},\] then what is the value of \[({{a}^{6}}+{{a}^{-6}})\]? |
A) 26910
B) 25800
C) 2400
D) 1390
Correct Answer: A
Solution :
| Given, \[\left( a+\frac{1}{a} \right)=4\sqrt{2}\] |
| On squaring both sides, we get |
| \[{{\left( a+\frac{1}{a} \right)}^{2}}={{(4\sqrt{2})}^{2}}\] |
| \[\Rightarrow \] \[{{(4\sqrt{2})}^{2}}={{a}^{2}}+\frac{1}{{{a}^{2}}}+2\] |
| \[[\because {{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab]\] |
| \[\Rightarrow \] \[{{a}^{2}}+\frac{1}{{{a}^{2}}}=32-2=30\] |
| Now, on cubing both sides, we get |
| \[{{\left( {{a}^{2}}+\frac{1}{{{a}^{2}}} \right)}^{3}}={{(30)}^{3}}\] |
| \[\Rightarrow \]\[{{a}^{6}}+\frac{1}{{{a}^{6}}}+3\cdot {{a}^{2}}\times \frac{1}{{{a}^{2}}}\left( {{a}^{2}}+\frac{1}{{{a}^{2}}} \right)=27000\] |
| \[[\because {{(a+b)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\,(a+b)]\] |
| \[\Rightarrow \] \[{{a}^{6}}+\frac{1}{{{a}^{6}}}+3\,\,(30)=27000\] |
| \[\Rightarrow \]\[{{a}^{6}}+\frac{1}{{{a}^{6}}}=27000-90\]\[\Rightarrow \]\[{{a}^{6}}+\frac{1}{{{a}^{6}}}=26910\] |
| \[\therefore \]\[{{a}^{6}}+{{a}^{-6}}=26910\] |
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